Electricity-And-Magnetism · Unit 2 · Video 6 · Interactive Practice

Drawing Vector Fields: Arrows, Lines and Grass Seeds

IKey Formulas

Formula	Name	What it says
$\vec E \propto \dfrac{x\,\hat\imath + (y+d)\,\hat\jmath}{[x^2+(y+d)^2]^{3/2}} - \dfrac{1}{3}\,\dfrac{x\,\hat\imath + (y-d)\,\hat\jmath}{[x^2+(y-d)^2]^{3/2}}$	The $+3$ / $-1$ pair in the plane $z = 0$	Each term is (vector from the charge) / distance $^3$ , so it falls off as $1/\text{distance}^2$
$\dfrac{d\vec r}{ds} = \hat E = \dfrac{\vec E}{\|\vec E\|}$	Field line	Tangent to the field at every point; lines never cross
$\vec v = \dfrac{\hat r}{r^2} = \dfrac{\vec r}{r^3} = \dfrac{x\,\hat\imath + y\,\hat\jmath}{(x^2+y^2)^{3/2}}$	Inverse-square radial field	Points away from the origin; $\|\vec v\| = 1/r^2$
$\vec v = \dfrac{3xy}{r^5}\,\hat\imath + \dfrac{2y^2 - x^2}{r^5}\,\hat\jmath$	Point dipole at the origin	$\|\vec v\| = \dfrac{\sqrt{x^2+4y^2}}{r^4} \propto \dfrac{1}{r^3}$

Key Insight: Field lines never cross, because at any point where $\vec E \neq \vec 0$ the field has exactly one direction. Lines meet only at the charges and at a zero of the field, which for the $+3$ / $-1$ pair sits at $y = (2+\sqrt3)\,d$ .

IIBetween the Field Lines

One field, three pictures, and through every point between the drawn lines there runs another field line.

Picture Lines leaving +3 12

IIIFaraday's Construction of a Field Line

Short straight steps along the local field build a polygon that converges to the field line as the step shrinks.

Step length h Progress 100%

IVThe Four Fields of the Worked Problem

An arrow grid shows direction; its length or shade can also encode $|\vec v|$ , the information that separates $\hat r$ from $\hat r / r^2$ .

Field Arrows

VQuiz Questions

Problem 1 · The Inverse-Square Field at a Point

Given: $\vec v = \hat r / r^2$ in the plane — find $\vec v$ at the point $(6, 8)$ .

✅ Correct!

\vec v = \vec r / r^3

with

r = 10

, so

|\vec v| = 1/100 = 1/r^2

❌ That is $\hat r$ alone. It has length

1

; the inverse-square field still has to be divided by

r^2 = 100

❌ That is $\vec r / r^2$ . Its length is

1/r = 1/10

. Since

\hat r = \vec r / r

, the field

\hat r / r^2

\vec r / r^3

❌ That is $\vec r / r^4$ . Its length is

1/r^3

; one power of

r

too many.

Show solution

At $(6, 8)$ the distance is $r = \sqrt{6^2 + 8^2} = 10$ . Since $\hat r = \vec r / r$ ,

\vec v = \frac{\hat r}{r^2} = \frac{\vec r}{r^3} = \frac{6\,\hat\imath + 8\,\hat\jmath}{1000} = 0.006\,\hat\imath + 0.008\,\hat\jmath.

Check: $|\vec v| = \sqrt{0.006^2 + 0.008^2} = 0.01 = 1/r^2$ . Compared with the video's point $(3, 4)$ , where $|\vec v| = 1/25$ , doubling the distance made the arrow a quarter as long.

Problem 2 · Where the Field Lines Meet

Given: the video's pair, $+3$ at $(0, -d)$ and $-1$ at $(0, d)$ , with $\vec E \propto \dfrac{\vec r - \vec r_+}{|\vec r - \vec r_+|^3} - \dfrac13\,\dfrac{\vec r - \vec r_-}{|\vec r - \vec r_-|^3}$ — find the point on the $y$ axis where $\vec E = \vec 0$ .

✅ Correct! Beyond the smaller charge the two terms point in opposite directions and their magnitudes match at

y = (2+\sqrt3)\,d

❌ The square root is missing. Each term falls off as

1/\text{distance}^2

, so equal magnitudes give

(y+d)^2 = 3(y-d)^2

, not

y + d = 3(y - d)

❌ That root is extraneous. Between the charges the

+3

term pushes up and the

-1

term pulls up, so the two add; at

y \approx 0.27\,d

the field is about

1.24

in these units, pointing up.

❌ Below the $+3$ charge the larger charge is also the nearer one, so its term always wins and the field never vanishes there.

Show solution

On the $y$ axis ( $x = 0$ ) both terms are vertical. Above the negative charge ( $y > d$ ) the $+3$ term points up and the $-1$ term points down, so

E_y \propto \frac{1}{(y+d)^2} - \frac{1}{3}\,\frac{1}{(y-d)^2} = 0 \;\Longrightarrow\; (y+d)^2 = 3\,(y-d)^2.

Taking the positive root (both $y + d$ and $y - d$ are positive here): $y + d = \sqrt3\,(y - d)$ , so

y = \frac{\sqrt3 + 1}{\sqrt3 - 1}\,d = (2 + \sqrt3)\,d \approx 3.73\,d.

The other root, $y = (2-\sqrt3)\,d$ , lies between the charges, where both terms point up and cannot cancel. Below $y = -d$ the $+3$ charge is both larger and nearer, so there is no zero there either. This is the single zero of the field: lines come into it from left and right and leave it up and down the $y$ axis.

Problem 3 · Reading the Dipole Field

Given: part (d), $\vec v = \dfrac{3xy}{r^5}\,\hat\imath + \dfrac{2y^2 - x^2}{r^5}\,\hat\jmath$ with $r = \sqrt{x^2 + y^2}$ — find $\vec v$ at $(3, 4)$ , and how $|\vec v|$ at $(6, 8)$ compares with $|\vec v|$ at $(3, 4)$ .

What is $\vec v(3, 4)$ ?

$|\vec v(6, 8)|\,/\,|\vec v(3, 4)| = \;?$

✅ Correct!

\vec v(3,4) = (36\,\hat\imath + 23\,\hat\jmath)/3125

, and the dipole field falls off as

1/r^3

, so doubling

r

divides

|\vec v|

8

❌ Check the components.

3xy = 36

and

2y^2 - x^2 = 32 - 9 = 23

, both over

r^5 = 5^5 = 3125

❌ The denominator is $r^5 = 3125$ , not

r^4 = 625

; it is the magnitude that carries

r^4

, as

\sqrt{x^2+4y^2}/r^4

❌ Use the magnitude formula.

|\vec v| = \sqrt{x^2 + 4y^2}/r^4

: the numerator doubles and

r^4

grows by

16

❌ That is the inverse-square rule of a single charge. The dipole's two halves nearly cancel far away, so it falls off one power faster, as

1/r^3

Show solution

At $(3, 4)$ : $r = 5$ , $r^5 = 3125$ , $3xy = 36$ , $2y^2 - x^2 = 2(16) - 9 = 23$ , so

\vec v(3, 4) = \frac{36}{3125}\,\hat\imath + \frac{23}{3125}\,\hat\jmath \approx 0.01152\,\hat\imath + 0.00736\,\hat\jmath.

Its magnitude: $\dfrac{\sqrt{36^2 + 23^2}}{3125} = \dfrac{\sqrt{1825}}{3125} = \dfrac{5\sqrt{73}}{3125} = \dfrac{\sqrt{73}}{625} \approx 0.01367$ , which matches $\sqrt{x^2 + 4y^2}/r^4 = \sqrt{9 + 64}/625$ .

At $(6, 8)$ : $r = 10$ and $|\vec v| = \sqrt{36 + 256}/10^4 = \sqrt{292}/10^4 = 2\sqrt{73}/10^4$ . The ratio is

\frac{2\sqrt{73}/10^4}{\sqrt{73}/625} = \frac{1250}{10^4} = \frac{1}{8} = \frac{1}{2^3}.

Problem 4 · Which Way Does the Line Run?

Given: the $+3$ / $-1$ pair with $d = 1$ — find the slope $dy/dx$ of the field line passing through the point $(1, 0)$ .

✅ Correct! The field line is tangent to

\vec E

, and

\vec E(1, 0) \propto (\tfrac{1}{3},\, \tfrac{2}{3})/\sqrt2

has slope

2

❌ That is the direction of the $+3$ term alone. The

-1

term also contributes at

(1, 0)

, pulling the field toward

(0, 1)

❌ The $-\tfrac13$ term was added with the wrong sign. It points toward the negative charge, i.e. along

-(1, -1)

, adding to

E_y

and subtracting from

E_x

❌ That would be a $+1$ / $-1$ pair. With equal magnitudes the

x

components cancel on the line

y = 0

; here the negative term carries the factor

\tfrac13

Show solution

From $(0, -1)$ to $(1, 0)$ the vector is $(1, 1)$ with length $\sqrt2$ ; from $(0, 1)$ it is $(1, -1)$ , also length $\sqrt2$ . So

\vec E \propto \frac{(1,\,1)}{2\sqrt2} - \frac{1}{3}\,\frac{(1,\,-1)}{2\sqrt2} = \frac{1}{2\sqrt2}\left(\tfrac{2}{3},\ \tfrac{4}{3}\right) \approx (0.2357,\ 0.4714).

A field line is tangent to $\vec E$ , so its slope is $\dfrac{dy}{dx} = \dfrac{E_y}{E_x} = \dfrac{4/3}{2/3} = 2$ : the line leaves $(1, 0)$ heading up and to the right at $\arctan 2 \approx 63.4\degree$ above the $+x$ axis.

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